4
$\begingroup$

The process $S$ is a geometric Brownian motion with an SDE: $dS_t = S_t(\sigma\, dB_t + \mu\, dt)$. I'm stuck evaluating $E(X_t)$ and $V(X_t)$, where $dX_t = t\,dS_t$.

$\endgroup$
4
$\begingroup$

Using Itô's Lemma, notice that: $$d(tS_t)=tdS_t+S_tdt=dX_t+S_tdt$$ Hence: $$X_t=tS_t-\int S_udu$$ Using independence of Brownian increments, $E(S_udW_u)=E(S_u)E(dW_u)=0$, and the chain rule for the 4th step: $$\begin{align} E(X_t)&=E\left(\int dX_u\right) \\ &=\int uE(dS_u) \\ &=\int u\mu E(S_u)du \\ &=S_0\int u\mu e^{\mu u}du \\ &=S_0\left(te^{\mu t}-\int e^{\mu u}du\right) \\ &=S_0\left(te^{\mu t}-\frac{1}{\mu}(e^{\mu t}-1)\right) \end{align} $$

[Note: my previous variance calculation was wrong, will fix it when available.]

$\endgroup$
  • $\begingroup$ Thanks. Quite to the point. Solving the equation for Xt, can I use ito lemma with f(x,t)=d((log s)/t)? $\endgroup$ – Charles Smith Oct 17 at 8:26
  • $\begingroup$ @CharlesSmith I have updated my answer. $\endgroup$ – Daneel Olivaw Oct 17 at 10:42
  • $\begingroup$ I am not sure whether $f(S_t,t)=\log (S_t)/t$ is the best option. $\endgroup$ – Daneel Olivaw Oct 17 at 10:46
  • $\begingroup$ @DaneelOlivaw I am pretty sure the plus signs in the last two expressions in the calculation of the expectation should be minus signs. Also, (you seem much more experienced so I may be totally off base here) I think that there may be an issue in calculating the Variance's first term since in the Variance the $\mathbb E [(\int u \mu S_u du)^2]$ term should appear as a double integral using two different dummy variables. So the $dt^2$ argument doesn't seem to work. $\endgroup$ – Slade Oct 17 at 13:20
  • $\begingroup$ @Slade correct on both counts, I did this too hastily. For variance, I was using independence of $S_u$ and $dW_u$ to separate the total variance into the sum of the "drift part" and the "diffusion part", but I messed it up as $S_u$ multiplies $dW_u$ so the two integrals are not independent. $\endgroup$ – Daneel Olivaw Oct 17 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.